Proof of Theorem 1, Chapter 12: Modules over Principal Ideal Domains

[quasar987]

[SOLVED] Algebra - modules

Homework Statement

I'm reading this proof from D&F and there's something I don't get. It is theorem 1 of chapter 12 on modules over principal ideal domains. The theorem is the following

"Let R be a ring and let M be a left R-module. Then,

(Ever nonempty set of submodules of M contains a maximal element under inclusion) ==> (Every submodule of M is finitely generated)"

The authors prove this assertion by letting N be a submodule of M and denoting S the collection of all finitely generated submodules of N. The hypothesis guarantees the existence of a maximal element N' of S, and we'd be happy if N'=N. So suppose N' is different from N and consider an element x in N\N'. Then the submodule of N generated by {x}uN' is finitely generated, thus violating the maximality of N'.

The Attempt at a Solution

How is this so? Notice that R is not assumed to have a 1. So there is nothing I can see that guarantees that N' is contained, let alone properly contained, in <{x}uN'>.

 

[lippyka]

Where am I going wrong?The key here is that R is a principal ideal domain, so we can use the fact that every ideal of R is generated by a single element. In particular, since N' is a finitely generated submodule of N, it follows that N' must be generated by some set S = {x_1,...,x_n} of elements in R. Then, the submodule of N generated by {x}uN' must also be finitely generated and thus must contain S, since S generates N'. This contradicts the maximality of N', as N' was assumed to be contained in N but not in <{x}uN'>.